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Let $W(t_1,\dotsc,t_n)$ a holomorphic function on some connected open set $U$ of $\mathbb C^n$. Let $\mathbf t^{(0)}$ a point of $U$.

Assume that there exists a cycle $\gamma$ in $\mathbb C^m$ and a rational function $F(\mathbf t, x_1,\dotsc, x_m)$ such that for all $\mathbf t$ in a neighbourhood of $\mathbf t^{(0)}$:

  1. $\mathbf x \mapsto F(\mathbf t, \mathbf x)$ is continuous on $\gamma$;
  2. $\displaystyle W(\mathbf t) = \oint_\gamma F(\mathbf t, \mathbf x)\mathrm d \mathbf x$.

Is it true that for all point $\mathbf t^{(1)}$ in $U$, there exists another (sum of) cycle $\gamma_1$ such that properties 1. and 2. are satisfies in a neighbourhood of $\mathbf t^{(1)}$ ?

For simple integrals, that is $m=1$, this is true, and not too hard to see: the poles of $F(\mathbf t, x)$ are points that move continuously with $\mathbf t$. It is easy to deform $\gamma$ so that is does not encounter these moving points. There is a singularity when a pole inside $\gamma$ collapses with a pole outside, but this can't happen if we stay in the domain of holomorphy of $W$. However, it seems harder for multiple integrals...

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Not an answer, but here's a reference that might be helpful: Tsikh, Multidimensional Residues and Their Applications (AMS, 1992) books.google.ca/books?id=SKCN1eN7hz0C –  Igor Khavkine Jan 18 '14 at 23:45
Thanks, I'll see if I can get that book. –  Lierre Jan 19 '14 at 0:59

1 Answer 1

up vote 2 down vote accepted

I'm not sure you're right about the $m=1$ case. Take $$F(t,x) = \frac{2x^3}{x^2-t} = \frac{x^2}{x-\sqrt{t}} + \frac{x^2}{x+\sqrt{t}}$$ take $t^0=1$ and take $\gamma$ to be a circle in the $x$-plane which encloses $1$ but not $-1$. Then $$\oint_{\gamma} F(x,t) dx = (2 \pi i) (\sqrt{t})^2 = (2 \pi i) t$$ where $\sqrt{t}$ means the square root of $t$ which lies in $\gamma$.

Now, the function $W(t) = (2 \pi i) t$ extends holomorphically to the whole $t$-plane. Consider $t^1$ to be the point $t=0$.

If I understand correctly, you are claiming that there is a cycle $\gamma^1$ so that $F(t,x)$ will be continuous on $t \times \gamma^1$ for any $t$ near $0$, and this cycle will have $\int_{\gamma_1} F(t,x) dx = (2 \pi i) t$.

But $F$ is discontinuous at $(0,0)$, so $\gamma_1$ must not pass through $0$. Thus, either $\gamma_1$ encloses $0$ or it doesn't. So, for $t$ near but not equal to $0$, either $\gamma_1$ encloses both poles of $F$ or neither one, and we get $\oint_{\gamma_1} F = 2 (2 \pi i) t$ or $0$.

In other words, I disagree with your statement "There is a singularity when a pole inside $\gamma$ collapses with a pole outside, but this can't happen if we stay in the domain of holomorphy of $W$." If I rig the pole residues carefully, I can make $W$ stay holomorphic even though the pieces it is built out of are becoming singular.

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In retrospect, an easier example is $F(t,x) = (2x)/(x^2-t) = 1/(x-\sqrt{t}) + 1/(x+\sqrt{t})$. –  David Speyer Jan 27 '14 at 18:17
Still seems impossible, with the same example. How do you get a contour $\gamma$ which, for $t$ in a punctured neighborhood of $0$, always has precisely one of $\pm \sqrt{t}$ inside $\gamma$, with $\pm \sqrt{t}$ never landing on $\gamma$ itself? –  David Speyer Jan 27 '14 at 22:22
Damn! You got this one ;) Too bad the answer is not positive. However, if you allow linear combination of cycle, then you can choose $\gamma_1$ to be $\frac 12$ times the unit circle. But I'm not sure this is a real objection. (And you right for the punctured disc, that was pointless ;) –  Lierre Jan 27 '14 at 22:22

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